Measures and Their Basic Properties — Examples

For the forward implication, suppose counting measure on $X$ is sigma-finite. By [L2], $X={\bigcup }_n{E}_n$ with $\mathrm{\#}(E_n)<+\mathrm{\infty }$, so every $E_n$ is finite by [L1] and $X$ is at most countable by [L3].

For the reverse implication when $X$ is finite, the constant cover $E_n=X$ has finite counting measure; when $X$ is countably infinite, a bijection $e:\mathbb{N}\to X$ gives finite initial sets $E_n=\{e(k):k<n\}$ with union $X$. The empty set is covered by the constant empty sequence.

Steps 1.1 and 1.2 prove the equivalence. Since $\mathbb{R}$ is uncountable by [L4], the forward implication shows that its counting measure is not sigma-finite.

For every measurable $E$, [L1] gives ${\delta }_{x_0}(E)=1$ if $x_0\in E$ and $0$ otherwise; in particular its values on $\mathrm{\varnothing }$ and $X$ are $0$ and $1$.

By [L2], $({\delta }_{x_0}{)}_F(A)={\delta }_{x_0}(A\cap F)$. If $x_0\in F$, this equals $1$ exactly when $x_0\in A$, so it is ${\delta }_{x_0}(A)$; if $x_0\mathrm{\notin }F$, it is always $0$.

Steps 1.1 and 1.2 verify the concentration and both restriction cases, including $F=\mathrm{\varnothing }$ and $F=X$.

By [L1], $\mu ={\sum }_k{2}^{-(k+1)}{\delta }_k$ is a measure, and evaluating it on $E$ gives exactly the displayed subseries because ${\delta }_k(E)$ is $1$ for $k\in E$ and $0$ otherwise.

The geometric-series formula with $r=1\mathrm{/}2$ gives $\mu (\mathbb{N})={\sum }_{k=0}^{\mathrm{\infty }}{2}^{-(k+1)}=(1\mathrm{/}2){\sum }_{k=0}^{\mathrm{\infty }}(1\mathrm{/}2{)}^k=1$.

The unshifted total is ${\sum }_{k=0}^{\mathrm{\infty }}{2}^{-k}=2$, whose first term at $k=0$ is $1$; thus those weights do not define a probability measure.

Steps 1.1 and 1.2 make $\mu$ a probability measure by [L2], and step 1.3 verifies the index-zero boundary and rules out the unshifted construction.

The countable-cocountable family $\mathcal{A}$ is a sigma-algebra: complements exchange its two classes, a countable union of countable members is countable by [L2], and a union containing a cocountable member is cocountable.

The whole space has measure $+\mathrm{\infty }$, while every measurable set of finite measure has measure $0$; hence $X$ contains no measurable subset of positive finite measure.

In a pairwise disjoint sequence from $\mathcal{A}$, at most one member is cocountable. If all members are countable, their union is countable by [L2] and both sides of countable additivity are $0$; if one is cocountable, the union is cocountable and both sides are $+\mathrm{\infty }$.

Steps 1.1 and 2.1 prove that $\mu$ is a measure, and step 1.2 violates the semifiniteness condition [L3].

The singleton $\{0\}$ is Borel because its complement is the union of the open rays $(-\mathrm{\infty },0)$ and $(0,+\mathrm{\infty })$; hence $Z:=\mathbb{R}\setminus \{0\}$ is Borel and ${\delta }_0(Z)=0$.

The binary-sequence injection [L5] gives an injection $f$ from a set of cardinality $\mathfrak{c}$ into $\mathbb{R}$; the map $S\mapsto f[S]$ then injects its power set into $\mathcal{P}(\mathbb{R})$. Hence [L4] gives $\mathrm{\mid }\mathcal{P}(\mathbb{R})\mathrm{\mid }\ge 2^{\mathfrak{c}}>\mathfrak{c}=\mathrm{\mid }\mathcal{B}(\mathbb{R})\mathrm{\mid }$, so some subset $S$ of $\mathbb{R}$ is not Borel.

If $0\mathrm{\notin }E$, then $E=\mathrm{\varnothing }\cup E$ with $E\subseteq Z$ is a completed representation and has completed value $0$ by [L6]; if $0\in E$, then $E=\{0\}\cup (E\setminus \{0\})$ with the second part contained in $Z$, so [L2] and [L6] give completed value $1$.

Thus every subset of $\mathbb{R}$ lies in the completion and the displayed formula holds.

For the non-Borel set $S$ from step 1.2, the set $S\setminus \{0\}$ is still non-Borel, since otherwise adjoining the Borel singleton $\{0\}$ when necessary would make $S$ Borel; it is a subset of the Borel null set $Z$ from step 1.1. Thus the original Borel Dirac space is not complete, while step 3.1 shows that its completion is the full power set.

By [L1], $\mu$ is a measure. For fixed $k$, the atom $2^{-j}$ lies in $(0,2^{-k})$ exactly when $j>k$, so $\mu (E_k)={\sum }_{j>k}{2}^{-(j+1)}$.

The intervals decrease. No $x\le 0$ belongs to any of them, and for $x>0$, convergence $2^{-k}\to 0$ supplies $k$ with $2^{-k}\le x$, so $x\mathrm{\notin }{E}_k$; hence their intersection, and therefore their limsup, is empty.

The tail geometric sum in step 1.1 is $2^{-(k+1)}$, and therefore ${\sum }_{k=0}^{\mathrm{\infty }}\mu (E_k)={\sum }_{k=0}^{\mathrm{\infty }}{2}^{-(k+1)}=1$. At $k=0$, this gives $\mu ((0,1))=1\mathrm{/}2$.

Borel-Cantelli applied using step 2.1 gives $\mu ({\mathrm{lim\hspace{0.17em}sup}}_k{E}_k)=0$, while step 1.2 independently identifies that limsup as $\mathrm{\varnothing }$.

The sequence decreases because $k+1\le n$ implies $k\le n$, and $E_0=\mathbb{N}$.

Every $E_k$ is infinite, since $n\mapsto k+n$ injects $\mathbb{N}$ into $E_k$, so $\mathrm{\#}(E_k)=+\mathrm{\infty }$.

The intersection is empty: a natural $n$ does not belong to $E_{n+1}$.

Hence $\mathrm{\#}({\bigcap }_k{E}_k)=0$ but ${\inf }_k\mathrm{\#}(E_k)=+\mathrm{\infty }$. The hypothesis of [L2] fails at every index, exactly as required.

The family $\mathcal{B}=\{B_n:n\in \mathbb{N}\}$ is a filter base: $B_0=\mathbb{N}$, no tail is empty, and $B_{\max (m,n)}\subseteq B_m\cap B_n$. Let $\mathcal{F}=⟨\mathcal{B}⟩$.

By [L2], extend $\mathcal{F}$ to an ultrafilter $\mathcal{U}$. It is free: if $\{m\}\in \mathcal{U}$, then $B_{m+1}\in \mathcal{F}\subseteq \mathcal{U}$, so closure under intersections would put $\mathrm{\varnothing }=\{m\}\cap B_{m+1}$ in the filter.

Define $q(A)=1$ when $A\in \mathcal{U}$ and $q(A)=0$ otherwise. Then $q(\mathrm{\varnothing })=0$ and $q(\mathbb{N})=1$.

If $A,B$ are disjoint, then $A\cup B\in \mathcal{U}$ exactly when one of $A,B$ lies in $\mathcal{U}$: the reverse implication is upward closure; for the forward implication, if $A\mathrm{\notin }\mathcal{U}$, then $A^c\in \mathcal{U}$ and $(A\cup B)\cap A^c=B$ lies in $\mathcal{U}$. Disjointness prevents both from lying in $\mathcal{U}$. Thus $q(A\cup B)=q(A)+q(B)$, so $q$ is finitely additive.

Every singleton has value $0$ because $\mathcal{U}$ is free, but their disjoint union is $\mathbb{N}$ and has value $1$. Hence countable additivity fails.

Steps 4.1 and 4.2 give a finitely additive zero-one probability that is not countably additive, refuting the proposed implication.

Give $\mu$ masses $(1\mathrm{/}2,0,0,1\mathrm{/}2)$ at $(NW,NE,SW,SE)$ and give $\nu$ masses $(0,1\mathrm{/}2,1\mathrm{/}2,0)$. Finite summation over subsets defines probability measures on $\mathcal{P}(X)$.

Intersecting rows with columns produces every singleton, so ${\sigma }_X(\mathcal{G})=\mathcal{P}(X)$; a singleton intersection is absent from $\mathcal{G}$, so $\mathcal{G}$ is not a pi-system. The constant sequence $X,X,\dots$ is an increasing finite-measure exhaustion lying in $\mathcal{G}$.

Every row and column contains exactly one $\mu$-atom and one $\nu$-atom of mass $1\mathrm{/}2$, and both measures give $X$ mass $1$, so the measures agree on $\mathcal{G}$.

The measures are unequal because $\mu (\{NW\})=1\mathrm{/}2$ and $\nu (\{NW\})=0$, despite steps 2.1 and 1.2. The family satisfies the generating and finite-measure exhaustion hypotheses and violates the pi-system hypothesis of [L3].